If $2^d-1\mid 2^n-1$ then $d\mid n$

Asked Oct 16 '23 at 14:28

Active Oct 16 '23 at 14:52

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It is well-known that if $d,n \in \mathbb{N}^*$ and $d \mid n$ then $2^d -1 \mid 2^n - 1$.

Is the reciprocal true ? With a Python program it shows that it may be true (didn't found any counterexample) but I haven't been able to prove it.

My attempt :

By contraposition, if $d \nmid n$ then exists $(q,r)\in\mathbb{N} \times \mathbb{N}^*$ such that $n=dq+r$. Then :

$$2^n-1 = 2^{dq}2^r -1$$

And I tried to watch what's happening modulo $2^d-1$ but it wasn't succesful.

If anybody knows what to do I would be grateful.

edited Oct 16 '23 at 14:52

Bill Dubuque

asked Oct 16 '23 at 14:28

LexLarn

2

See this question – lulu Oct 16 '23 at 14:32
Wait, if $2^d - 1 \mid 2^n - 1$ we have $\mathrm{gcd}(2^d - 1, 2^n-1) = 2^d-1$, but also we have $\mathrm{gcd}(2^d - 1, 2^n-1) = 2^{\mathrm{gcd}(d,n)} - 1$. So $d = \mathrm{gcd}(d,n)$, and we can conclude that $d \mid n$ ? – LexLarn Oct 16 '23 at 14:35
Yes, that's the point. – lulu Oct 16 '23 at 14:39

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